Question

Answer questions 16 – 17 based on the following. A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information:

	Today	Five Years Ago
x_bar	82	88
σ^2	112.5	54
n	45	36

16. The 95% confidence interval for the difference between the two population means (Today – Five Years Ago) is

a. -9.92 to -2.08

b. -3.08 to 3.92

c. -13.84 to -1.16

d. -24.77 to 12.23

e. 2.08 to 9.92

17. The statistics teacher wishes to test, using a two-tailed approach, the hypothesis of no difference between the population mean scores using a 5% level of significance. The p-value associated with this test is: When computing two-tailed p-values, remember to use the 2p approach!

a. 0.0013

b. 0.0026

c. 0.4987

d. 0.9987

Answer 1

16)
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(112.49996356/45 + 54.00045225/36)
sp = 2

Given CI level is 0.95, hence α = 1 - 0.95 = 0.05                  
α/2 = 0.05/2 = 0.025, zc = z(α/2, df) = 1.96                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 1.96 * 2                  
ME = 3.92                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (82 - 88 - 1.96 * 2 , 82 - 88 - 1.96 * 2                  
CI = (-9.92 , -2.08)

17)

sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(112.49996356/45 + 54.00045225/36)
sp = 2

Test statistic,
z = (x1bar - x2bar)/sp
z = (82 - 88)/2
z = -3

P-value Approach
P-value = 0.0026